Quantitative Aptitude PYQs

Follow BODMAS rule to solve this question, as per the order given below,

Step-1- Parts of an equation enclosed in 'Brackets' must be solved first, and in the bracket,

Step-2- Any mathematical 'Of' or 'Exponent' must be solved next,

Step-3- Next, the parts of the equation that contain 'Division' and 'Multiplication' are calculated,

Step-4- Last but not least, the parts of the equation that contain 'Addition' and 'Subtraction' should be calculated.

Given expression is,

 [(368)² ÷ (23)² × 15] ÷ 12 = ?

⇒ [(368 ÷ 23)² × 15] ÷ 12 = ?

Applying BODMAS Rule;

⇒ [16² × 15] ÷ 12 = ?

⇒ [256 × 15] ÷ 12 = ?

⇒ 3840 ÷ 12 = ?

∴ ? = 320

Concept Used:

Follow the BODMAS rule as given in the below table:

Calculation:

28 ÷ (√144 × 17 ÷ 51) + (?)2 = 848

Applying BODMAS Rule;

⇒ 28 ÷ [12 × (1/3)] + (?)2 = 848

⇒ 28 ÷ 4 + (?)2 = 848

⇒ 7 + (?)2 = 848

⇒ (?)2 = 848 - 7 = 841

⇒ (?) = √841 = 29

∴ ? = 29

Follow BODMAS rule to solve this question, as per the order given below,

Step-1- Parts of an equation enclosed in 'Brackets' must be solved first, and in the bracket,

Step-2- Any mathematical 'Of' or 'Exponent' must be solved next,

Step-3- Next, the parts of the equation that contain 'Division' and 'Multiplication' are calculated,

Step-4- Last but not least, the parts of the equation that contain 'Addition' and 'Subtraction' should be calculated.

Given expression is,

(12/5)² × (37.5 × 56) ÷ (984/41) = ?

Applying BODMAS Rule;

⇒ (144/25) × 2100 ÷ 24 = ?

⇒ (144/25) × 87.5 = ?

∴ ? = 504

Follow the BODMAS rule to solve the question,

435 ÷ 7.5 – 40% of 130 = ?

Any mathematical 'Of' or 'Exponent' must be solved,

⇒ 435 ÷7.5 – [(40/100) × 130] = ?

The parts of the equation that contain 'Division' and 'Multiplication' are calculated,

⇒ (435 ÷ 7.5) – 52 = ?

The parts of the equation that contain 'Addition' and 'Subtraction' should be calculated,

⇒ 58 – 52 = ?

∴ ? = 6

Given:

858 ÷ 39 × 74 – 1500 = (2)?

Concept Used:

Calculation:

858 ÷ 39 × 74 – 1500 = (2)^?

By using the BODMAS rule,

⇒ (858 ÷ 39) × 74 – 1500 = 2^?

⇒ (22 × 74) – 1500 = 2^?

⇒ 1628 – 1500 = 2^?

⇒ 128 = 2^?

⇒ 2⁷ = 2^?

∴ ? = 7

Follow the BODMAS rule to solve this question, as per the order given below, 

Step-1: Parts of an equation enclosed in 'Brackets' must be solved first and in the bracket, 

Step-2: Any mathematical 'Of' or 'Exponent' must be solved next, 

Step-3: Next, the parts of the equation that contain 'Division' and 'Multiplication' are calculated, 

Step-4: Last but not least, the parts of the equation that contain 'Addition' and 'Subtraction' should be calculated 

120 ÷ 5 × (3)³ ÷ 9 = (?) ÷ [8 ÷ (?)]

⇒ 120 ÷ 5 × (3)3 ÷ 9 = (?) ÷ (8/?)

⇒ 120 ÷ 5 × (3)³ ÷ 9 = (?) × (?)/8

⇒ (?)² = 24 × 3 × 8

⇒ (?) = 24

Follow BODMAS (Brackets, Of, Division, Multiplication, Addition, Subtraction) rule to solve this question, as per the order given below,

Step-1- Parts of an equation enclosed in 'Brackets' must be solved inside the bracket first,

Step-2-  Then any mathematical 'Of' or 'Exponent' must be solved,

Step-3- Next, the parts of the equation that contain 'Division' and 'Multiplication' are calculated,

Step-4- Last but not least, the parts of the equation that contain 'Addition' and 'Subtraction' should be calculated.

Given expression is,

⇒ 999 ÷ 3 – 165 = 75 + 111 - ?

⇒ 333 – 165 = 186 - ?

⇒ 168 = 186 - ?

⇒ ? = 186 – 168

⇒ ? = 18

\(\sqrt {676} \times \frac{{67}}{{100}} \div \frac{1}{{100}} = ;? + ;577\)

\(\Rightarrow 26 \times \frac{{67}}{{100}} \times 100 = ;?; + ;577\)

⇒ 1742 = ? + 577

⇒ ? = 1742 – 577

⇒ ? = 1165

Given

(?)2 ÷ 4 × 5 = 180

⇒ (?)2 ÷ 4 = 36

⇒ (?)² = 36 × 4 

∴ ? = 6 × 2 = 12.

GIVEN:

In a milk tank the quantity of milk is 40% more than water

31 litres of water is added to this mixture

The final ratio of milk and water = 5 : 8

CALCULATION:

Initial ratio of milk and water = 7 : 5

Let the initial quantity of milk and water be 7x litres and 5x litres respectively.

Final ratio of milk and water = 5 : 8

According to question,

(5x + 31)/7x = 8/5

⇒ 56x = 25x + 155

⇒ 31x = 155

⇒ x = 5

⇒ 7x = (7 × 5) litres

∴ Initial quantity of milk = 35 litres

Given:

If the radius of a sphere increases by 5 cm then the surface area increases by 726 cm²

Formula Used:

The surface area of a sphere = 4π R²

Where, R = radius of the sphere

Calculation:

Let, the initial radius of the sphere = R cm

After the radius is increased by 5 cm, the new surface area becomes = 4π (R + 5) cm²

According to the question,

4π [(R + 5)² - R²] = 726

⇒ 4π [R² + 10R + 25 - R²] = 726

⇒ 4 × (22/7) × [10R + 25] = 726

⇒ 4/7 × [10R + 25] = 33

⇒ 40R + 100 = 231

⇒ R = 131/40 = 3.275 cm

∴ The initial radius of the sphere was 3.275 cm

Calculation:

k = |Online (Thu) − Offline (Tue)|

Required value = k + 5

⇒ k = |25 − 54|

⇒ k = 29

⇒ k + 5 = 29 + 5

⇒ = 34

∴ The value of (k + 5) = 34.

Calculations:

⇒ Difference = 54 − 40 = 14

⇒ % more = 14 ÷ 40 × 100

⇒ % more = 35%

∴ Offline orders on Tuesday are 35% more than offline orders on Wednesday.

Calculations:

⇒ Offline (Sunday) = 30 × 1.40

⇒ Offline (Sunday) = 42

⇒ 6 parts = 42

⇒ 1 part = 42 ÷ 6 = 7

⇒ Online (Sunday) = 7 parts = 7 × 7 = 49

∴ Number of online orders received on Sunday = 49.

GIVEN:

U_s = Upstream speed of a boat

D_s  = Downstream speed of a boat

⇒ U_s : D_s = 3x : 5x      ----(1)

FORMULA USED:

1. Downstream speed = (Speed of a boat in still water + Speed of current)/2

2. Upstream speed = (Speed of a boat in still water - Speed of current)/2

ASSUMPTION:

Let the speed of boat in still water and the speed of stream be (u) and (v) km/hr

CALCULATION:

⇒ u + v = 5x      ----(2)

⇒ u - v = 3x      ----(3)

Solving the above equations we get,

⇒ Speed of stream = v = 1x

⇒ Speed of boat in still water = u = 4x

⇒ v : u = 1 : 4

Given:

An amount becomes Rs. 1800 after 2 years and Rs. 1890 after 3 years at CI

Concept Used:

Simple Interest (SI) = (P × R × T)/100

Compound Interest (CI) = P × (1 + R/100)T - P

Where, P = principal amount, R = rate of interest, and T = time duration

Calculation:

Interest earned on Rs. 1800 in (3 - 2) = 1 year is = 1890 - 1800 = Rs. 90

Here, P = Rs. 1800, SI (as it is for 1 year) = Rs. 90, and T = 1 year

So, R = (90 × 100)/1800 = 5%

Let, the actual principal amount  = A

According to the question,

⇒ A × (1 + 5/100)² = 1800

⇒ A × 441/400 = 1800

⇒ A = (1800 × 400)/441 ≈ Rs. 1633

∴ The correct answer is option 3

Important PointsIf time duration (T) = 1 year, and the rate of interest is per annum, then the interest earned for SI or, CI will be equal. That is why the formula of SI is used to calculate the Rate (R), although the question is about CI.

GIVEN:

Ram's age = 37.5% of Shiv's age

Ram : Shiv (6 years hence) = 9 : 19

EXPLANATION:

37.5% = 3/8

Let Ram's age be 3x and Shiv's age be 8x.

A.T.Q

(3x + 6)/(8x + 6) = 9/19

57x + 114 = 72x + 54

15 x = 60 

x = 4

Sum = 3x + 8x = 11x

11x = 11 × 4 = 44 yrs

∴ The sum is 44 years.

Given:

Pipe X fills tank in 10 hours

Pipe Y fills tank in 15 hours

Formula used:

Work = LCM of time taken

Efficiency = Work ÷ Time

Time taken together = Total work ÷ (Efficiency of X + Efficiency of Y)

Calculations:

⇒ LCM of 10 and 15 = 30 units

⇒ X’s 1 hour work = 30 ÷ 10 = 3 units

⇒ Y’s 1 hour work = 30 ÷ 15 = 2 units

⇒ Total 1 hour work = 3 + 2 = 5 units

⇒ Time taken together = 30 ÷ 5 = 6 hours

∴ Both pipes together will fill the cistern in 6 hours.